Energy Budget for a Rocket in Free Space: Recall the analysis of the rocket engine in Project I, Flight of the Rocket. Although the momentum of the rocket and its exhaust is conserved (constant), their energy is not; energy must be supplied by the combustion of fuel. WARNING: This is not the same as the energy requirement of the jet airplane in Project II, Nonstop Service to Anywhere. Using the same approach we used to derive the First and Second Rocket Equation ---- sequential “snapshots” of the rocket---obtain an expression for the total kinetic energy gained by rocket and exhaust during the engine burn, in terms of the initial and final rocket masses Mi and Mf , the nozzle speed U of the exhaust, etc. This is the energy that must be supplied by combustion. HINTS: You may need to use the First Rocket Equation in your calculation. You may also use the fact that squares and products of differentials are negligible compared to the differentials themselves.
To derive an expression for the total kinetic energy gained by the rocket and exhaust during the engine burn, we can use the same approach as the First and Second Rocket Equations.
Let's consider a small time interval dt during which the rocket burns its fuel. At the beginning of this interval, the rocket has an initial mass Mi (including the remaining fuel) and a velocity vi. And at the end of this interval, the rocket has a final mass Mf and a velocity vf.
During this time interval, the rocket's mass decreases by dM (dM = Mi - Mf), and the velocity of the exhaust gas relative to the rocket is U (nozzle speed). The mass of the exhaust expelled during this time interval is dM and its velocity is U.
During the time interval dt, the change in the rocket's kinetic energy is given by:
dKE = (1/2) Mi vi^2 - (1/2) Mf vf^2
The change in the kinetic energy of the exhaust is given by:
dKE_exhaust = (1/2) dM U^2
Since the momentum of the rocket and its exhaust is conserved, we can equate the total momentum before and after the burn:
Mi * vi = Mf * vf + dM * U
Now, multiplying the equation by vf:
Mi * vi * vf = (Mf * vf^2) + (dM * vf * U)
Rearranging the equation:
(Mi * vi^2) - (Mf * vf^2) = (dM * vf * U)
Substituting this result back into the change in kinetic energy equation:
dKE = (dM * vf * U) + (1/2) dM * U^2
Now, summing the change in kinetic energy of the rocket and the exhaust:
Total dKE = dKE + dKE_exhaust
= (dM * vf * U) + (1/2) dM * U^2 + (1/2) dM * U^2
= (dM * vf * U) + dM * U^2
Integrating the total change in kinetic energy expression over the entire burn time:
ΔKE = ∫ [(dM * vf * U) + dM * U^2] from initial mass Mi to final mass Mf
Breaking down this integral:
ΔKE = ∫ (dM * vf * U) dM + ∫ dM * U^2 from Mi to Mf
Using the First Rocket Equation, which states that dM = U * dT / Ve (Ve is the exhaust velocity relative to the rocket):
ΔKE = U * ΔT * vf + U^2 * ∫ dM / Ve from Mi to Mf
Integrating the right term:
∫ dM / Ve = ∫ (Mi - Mf) / Ve = (Mi - Mf) / Ve
Substituting back into the previous equation:
ΔKE = U * ΔT * vf + U^2 * (Mi - Mf) / Ve
Finally, multiplying by the remaining mass flow rate (dm/dT):
Energy supplied by combustion = dm/dT * ΔKE
Therefore, the expression for the total kinetic energy gained by the rocket and exhaust during the engine burn, in terms of the initial and final rocket masses Mi and Mf, the nozzle speed U of the exhaust, and the remaining mass flow rate (dm/dT), is:
Energy supplied by combustion = (dm/dT) * [U * ΔT * vf + U^2 * (Mi - Mf) / Ve]